Let \(v(x,t)=\partial _tu(x,t)\) . Then by the definition of weak solution and Lemma REF  (ii), we have
\(\int _0^T\int _{\mathbb {R}^n}\partial _tv(x,t)\varphi (x,t)dxdt+\int _0^T\int _{\mathbb {R}^n}A\nabla v\cdot \nabla \varphi =0,\) 
for all \(\varphi \in L^2\left([0,T],W^{1,2}(\mathbb {R}^n)\right)\)  with \(\operatorname{supp}\varphi \subset \mathbb {R}^n\times (0,T]\) . By considering \(v^{\pm }\)  we can assume \(v\ge 0\) , and that
\(\int _0^T\int _{\mathbb {R}^n}\partial _tv(x,t)\varphi (x,t)dxdt+\int _0^T\int _{\mathbb {R}^n}A\nabla v\cdot \nabla \varphi \le 0,\) 
for all \(\varphi \in L^2\left([0,T],W^{1,2}(\mathbb {R}^n)\right)\)  with \(\operatorname{supp}\varphi \subset \mathbb {R}^n\times (0,T]\)  and \(\varphi \ge 0\)  a.e.
Now for any \(0\le s\le 1\) , define
\(Q_s=(1+s)Q, \quad I_s=((1-s)R_0^2,(2R_0)^2], \quad \mbox{and}\quad C_s=Q_s\times I_s.\) 
Fix \(l\in \mathbb {N}\) , define \(q_l=2k_0^l\) , where \(k_0=\frac{n+2}{n}\) . Note that \(q_0=2\) .
Furthermore, for any fixed
\(\frac{4}{3}\frac{1}{2^{l+2}}\le s_0<s_1\le \frac{3}{2}\frac{1}{2^{l+2}},\) 
choose \(\Psi _{s_0,s_1}\in C_0^2(C_{\frac{s_0+s_1}{2}})\)  and \(\widetilde{\Psi }_{s_0,s_1}\in C_0^2(C_{s_1})\)  such that \(\Psi _{s_0,s_1}=1\)  in \(C_{s_0}\) , \(\widetilde{\Psi }_{s_0,s_1}=1\)  in \(C_{\frac{s_0+s_1}{2}}\) ,
\(0\le \Psi _{s_0,s_1},\widetilde{\Psi }_{s_0,s_1}\le 1\) , and
\(\left|\nabla \Psi _{s_0,s_1}\right|^2+\left|\partial _t\Psi _{s_0,s_1}\right|+\left|\nabla \widetilde{\Psi }_{s_0,s_1}\right|^2+\left|\partial _t\widetilde{\Psi }_{s_0,s_1}\right|\lesssim \frac{R_0^{-2}}{(s_1-s_0)^2}.\) 
We omit the subscript \(s_0,s_1\)  in \(\Psi _{s_0,s_1}\)  and \(\widetilde{\Psi }_{s_0,s_1}\)  from now on.
Let \(t\in I_{s_0}\) . Recalling (REF ), one can take \(\varphi =v^{q_l-1}\Psi ^2\)  as a test function. Then (REF ) gives
\(\int _0^t\int _{\mathbb {R}^n}\partial _tv v^{q_l-1}\Psi ^2+\int _0^t\int _{\mathbb {R}^n}A\nabla v\cdot \nabla (v^{q_l-1}\Psi ^2)\le 0.\) 
For the first term, integration by parts gives
\(\int _0^t\int _{\mathbb {R}^n}\partial _tv v^{q_l-1}\Psi ^2&=\frac{1}{q_l}\int _{\mathbb {R}^n}v^{q_l}(x,t)\Psi ^2(x,t)dx-\frac{1}{q_l}\int _0^t\int _{\mathbb {R}^n}v^{q_l}\partial _t(\Psi ^2)\\&\ge \frac{1}{q_l}\int _{Q_{s_0}}v^{q_l}(x,t)dx-\frac{CR_0^{-2}}{q_l(s_1-s_0)^2}\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}.\) 
The second term in (REF ) is split, as usual, corresponding to the symmetric and antisymmetric part of \(A\) . Working with \(A^s\) , we estimate
\(\quad \int _0^t\int _{\mathbb {R}^n}A^s\nabla v\cdot \nabla (v^{q_l-1}\Psi ^2)\\=\frac{4(q_l-1)}{q_l^2}\int _0^t\int _{\mathbb {R}^n}A^s\nabla (v^{\frac{q_l}{2}})\cdot \nabla (v^{\frac{q_l}{2}})\Psi ^2\\+\frac{4}{q_l}\int _0^t\int _{\mathbb {R}^n}A^s\nabla (v^{\frac{q_l}{2}})\cdot \nabla \Psi v^{\frac{q_l}{2}}\Psi \\\ge \frac{2\lambda _0(q_l-1)}{q_l^2}\int _0^t\int _{Q_{s_0}}\left|\nabla v^{\frac{q_l}{2}}\right|^2dxdt-\frac{C(n,\lambda _0)R_0^{-2}}{q_l(s_1-s_0)^2}\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}.\) 
Turning to \(A^a\) , note that \(A^a\nabla v\cdot \nabla v \,\Psi ^2=0\)  due to anti-symmetry, so that
\(\int _0^t\int _{\mathbb {R}^n}A^a\nabla v\cdot \nabla (v^{q_l-1}\Psi ^2)=\int _0^t\int _{\mathbb {R}^n}A^a\nabla v\cdot \nabla (\Psi ^2)v^{q_l-1}\\=\frac{1}{q_l}\int _0^t\int _{\mathbb {R}^n}A^s\nabla (v^{q_l})\cdot \nabla (\Psi ^2)=\frac{1}{q_l}\int _0^t\int _{\mathbb {R}^n}A^s\nabla (v^{\frac{q_l}{2}}\widetilde{\Psi }v^{\frac{q_l}{2}}\widetilde{\Psi })\cdot \nabla (\Psi ^2).\) 
By Proposition REF ,
\(\left|\int _0^t\int _{\mathbb {R}^n}A^a\nabla v\cdot \nabla (v^{q_l-1}\Psi ^2)\right|\\\le \frac{C_n\Lambda _0}{q_l}\int _0^t\left\Vert \nabla \Psi \right\Vert _{L^{\infty }(\mathbb {R}^n)}\left\Vert v^{\frac{q_l}{2}}\widetilde{\Psi }\right\Vert _{L^2(\mathbb {R}^n)}\left\Vert \nabla \left(v^{\frac{q_l}{2}}\widetilde{\Psi }\right)\right\Vert _{L^2(\mathbb {R}^n)}dt\\\le \frac{C_{\theta }\Lambda _0}{q_l}\frac{R_0^{-2}}{(s_1-s_0)^2}\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt+\frac{\theta }{q_l}\int _{C_{s_1}}\left|\nabla v^{\frac{q_l}{2}}\right|^2.\) 
Combining these estimates with (REF ), we have
\(\int _{Q_{s_0}}v^{q_l}(x,t)dx+\int _0^t\int _{Q_{s_0}}\left|\nabla v^{\frac{q_l}{2}}\right|^2\\\le \frac{CR_0^{-2}}{(s_1-s_0)^2}\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt+C\theta \int _{C_{s_1}}\left|\nabla \left(v^{\frac{q_l}{2}}\right)\right|^2dxdt,\) 
where \(C=C(n,\lambda _0,\Lambda _0,\theta )\) .
Choosing \(\theta \)  to be sufficiently small, and then taking supremum in \(t\in I_{s_0}\) , we obtain
\(\sup _{t\in I_{s_0}}\int _{Q_{s_0}}v^{q_l}(x,t)dx+\int _{C_{s_0}}\left|\nabla v^{\frac{q_l}{2}}\right|^2dxdt\\\le \frac{CR_0^{-2}}{(s_1-s_0)^2}\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt+\frac{1}{2}\int _{C_{s_1}}\left|\nabla \left(v^{\frac{q_l}{2}}\right)\right|^2dxdt,\) 
which implies
\(\sup _{t\in I_{\frac{4}{3}\frac{1}{2^{l+2}}}}\int _{Q_{\frac{4}{3}\frac{1}{2^{l+2}}}}v^{q_l}(x,t)dx+\int _{C_{\frac{4}{3}\frac{1}{2^{l+2}}}}\left|\nabla v^{\frac{q_l}{2}}\right|^2dxdt\\\le C(n,\lambda _0,\Lambda _0)R_0^{-2}4^l\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt\) 
by Lemma REF .
Let us insert a cut-off function \(\Psi _l(x,t)\in C_0^2(C_{\frac{4}{3}\frac{1}{2^{l+2}}})\)  into (REF ) so that we can use an embedding theorem. As usual, \(\Psi _l\)  satisfies \(0\le \Psi _l\le 1\) , \(\Psi _l=1\)  in \(C_{\frac{1}{2^{l+2}}}\) , and
\(\left|\nabla \Psi _l\right|^2+\left|\partial _t\Psi _l\right|\lesssim R_0^{-2}4^l.\) 
Then we have
\(\sup _{t\in I_{\frac{4}{3}\frac{1}{2^{l+2}}}}\int _{Q_{\frac{4}{3}\frac{1}{2^{l+2}}}}v^{q_l}(x,t)\Psi _l(x,t)dx+\int _{C_{\frac{4}{3}\frac{1}{2^{l+2}}}}\left|\nabla (v^{\frac{q_l}{2}}\Psi _l)\right|^2dxdt\\\le CR_0^{-2}4^l\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt.\) 
Now, by a well-known embedding (see e.g. [1]} Theorem 6.9), we have
\(\int _{C_{\frac{1}{2^{l+2}}}}v^{q_lk_0}\le \int _{C_{\frac{4}{3}\frac{1}{2^{l+2}}}}(v^{\frac{q_l}{2}}\Psi _l)^{2k_0}\\\le \sup _{t\in I_{\frac{4}{3}\frac{1}{2^{l+2}}}}\left(\int _{Q_{\frac{4}{3}\frac{1}{2^{l+2}}}}(v^{\frac{q_l}{2}}\Psi _l)^2(x,t)dx\right)^{2/n}\int _{C_{\frac{4}{3}\frac{1}{2^{l+2}}}}\left|\nabla (v^{\frac{q_l}{2}}\Psi _l)\right|^2dxdt\\\le \left(\sup _{t\in I_{\frac{4}{3}\frac{1}{2^{l+2}}}}\int _{Q_{\frac{4}{3}\frac{1}{2^{l+2}}}}(v^{\frac{q_l}{2}}\Psi _l)^2(x,t)dx+\int _{C_{\frac{4}{3}\frac{1}{2^{l+2}}}}\left|\nabla (v^{\frac{q_l}{2}}\Psi _l)\right|^2dxdt\right)^{k_0}\\\le C\left(R_0^{-2}4^l\right)^{k_0}\left(\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt\right)^{k_0}.\) 
Therefore, for all \(l\in \mathbb {N}\) ,
\(\left(\int _{C_{\frac{1}{2^{l+2}}}}v^{q_{l+1}}dxdt\right)^{\frac{1}{q_{l+1}}}\le C^{\frac{1}{q_{l+1}}}(R_0^{-2}4^l)^{\frac{1}{q_l}}\left(\int _{C_{\frac{1}{2^{l+1}}}}v^{q_l}dxdt\right)^{\frac{1}{q_l}}.\) 
Then (REF ) follows from iteration and letting \(l\)  go to infinity.
